equivalence class of equinumerous sets is not a set

Recall that two sets are equinumerous iff there is a bijection between them.

Proposition 1.

Let A be a non-empty set, and E⁢(A) the class of all sets equinumerous to A. Then E⁢(A) is a proper class.

Proof.

E⁢(A)≠∅ since A is in E⁢(A). Since A≠∅, pick an elementa∈A, and let B=A-{a}. Then C:={y∣y⁢ is a set, and ⁢y∉B} is a proper class, for otherwise C∪B would be the “set” of all sets, which is impossible. For each y in C, the set F⁢(y):=B∪{y} is in one-to-one correspondence with A, with the bijection f:F⁢(y)→A given by f⁢(x)=x if x∈B, and f⁢(y)=a. Therefore E⁢(A) contains F⁢(y) for every y in the proper class C. Furthermore, since F⁢(y1)≠F⁢(y2) whenever y1≠y2, we have that E⁢(A) is a proper class as a result.
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Remark. In the proof above, one can think of F as a class function from C to E⁢(A), taking every y∈C into F⁢(y). This function is one-to-one, so C embeds in E⁢(A), and hence E⁢(A) is a proper class.